Question

Find the regression​ equation, letting the diameter be the predictor​ (x) variable. Find the best predicted circumference of a marble with a diameter of 1.5 cm. How does the result compare to the actual circumference of 4.7 ​cm? Use a significance level of 0.05.

	Baseball	Basketball	Golf	Soccer	Tennis	​Ping-Pong	Volleyball
Diameter	7.4	24.1	4.3	21.5	7.1	4.0	21.2
Circumference	23.2	75.7	13.5	67.5	22.3	12.6	66.6

The regression equation is

​(Round to five decimal places as​ needed.)

The best predicted circumference for a diameter of 1.5 cm is nothing cm.

​(Round to one decimal place as​ needed.)

How does the result compare to the actual circumference of 4.7 ​cm?

A.Even though 1.5 cm is beyond the scope of the sample​ diameters, the predicted value yields the actual circumference.

B.Since 1.5 cm is within the scope of the sample​ diameters, the predicted value yields the actual circumference.

C.Since 1.5 cm is beyond the scope of the sample​ diameters, the predicted value yields a very different circumference.

D.Even though 1.5 cm is within the scope of the sample​ diameters, the predicted value yields a very different circumference.

Answer 1